If $$I_{m,n} = \int_0^1 x^{m-1}(1-x)^{n-1}dx$$, for $$m, n \geq 1$$, and $$\int_0^1 \frac{x^{m-1} + x^{n-1}}{(1+x)^{m+n}} dx = \alpha I_{m,n}$$, $$\alpha \in R$$, then $$\alpha$$ equals ______.

We have $$I_{m,n} = \int_0^1 x^{m-1}(1-x)^{n-1}\,dx$$ (the Beta function $$B(m,n)$$), and we need to find $$\alpha$$ such that $$\int_0^1 \frac{x^{m-1} + x^{n-1}}{(1+x)^{m+n}}\,dx = \alpha\, I_{m,n}$$.

Consider $$J_1 = \int_0^1 \frac{x^{m-1}}{(1+x)^{m+n}}\,dx$$. Substitute $$u = \frac{x}{1+x}$$, so $$x = \frac{u}{1-u}$$, $$dx = \frac{1}{(1-u)^2}\,du$$, and $$1+x = \frac{1}{1-u}$$. When $$x = 0$$, $$u = 0$$; when $$x = 1$$, $$u = \frac{1}{2}$$.

Then $$J_1 = \int_0^{1/2} \frac{u^{m-1}}{(1-u)^{m-1}} \cdot (1-u)^{m+n} \cdot \frac{1}{(1-u)^2}\,du = \int_0^{1/2} u^{m-1}(1-u)^{n-1}\,du$$.

Similarly, $$J_2 = \int_0^1 \frac{x^{n-1}}{(1+x)^{m+n}}\,dx = \int_0^{1/2} u^{n-1}(1-u)^{m-1}\,du$$.

In $$J_2$$, substitute $$u = 1 - v$$: $$J_2 = \int_{1/2}^{1} (1-v)^{n-1} v^{m-1}\,dv = \int_{1/2}^1 v^{m-1}(1-v)^{n-1}\,dv$$.

Therefore, $$J_1 + J_2 = \int_0^{1/2} u^{m-1}(1-u)^{n-1}\,du + \int_{1/2}^1 u^{m-1}(1-u)^{n-1}\,du = \int_0^1 u^{m-1}(1-u)^{n-1}\,du = I_{m,n}$$.

Hence $$\alpha = 1$$.